Question

A basketball player throws a basketball m = 1 kg straight up with an initial speed of v0 = 5.5 m/s. The ball leaves his hand at shoulder height h0 = 2.15 m. Let gravitational potential energy be zero at ground level.Give the total mechanical energy of the ball E in terms of maximum height hm it reaches, the mass m, and the gravitational acceleration g. What is the height, hm in meters?

Answer 1

Using conservation of energy:

Let:

E1 = Energy right before the ball leaves his hand

E2 = Energy when the ball is in the air

$\begin{gathered} E1=E2 \\ \frac{1}{2}mv1^2+mgh1=\frac{1}{2}mv2^2+mgh2 \end{gathered}$

Where:

h2 = Maximum height

Solve for h2:

$\begin{gathered} v1^2+2gh1=v2^2+2gh2 \\ h2=\frac{v1^2+2gh1-v2^2}{2g} \\ so: \\ h2=\frac{5.5^2+2(9.8)(2.15)-0}{2(9.8)} \\ h2=\frac{72.39}{2(9.8)} \\ h2=3.6933 \end{gathered}$

The maximum height will be:

3.6933m